Let $M$ be a matrix made up of two diagonal blocks: $M = \begin{pmatrix} A & 0 \\ 0 & D\\ \end{pmatrix}$ Prove that $M$ is diagonalizable if and only if A and D are diagonalizable.
I know I'd have to use the minimal polynomial theorem, I just learnt it with one diagonal matrix in the diagonal, so for some reason I can't think of a way to implement it here
Is the minimal polynomial equation $M_{min} = A_{min} * D_{min}$ ? And after that can I say $A_{min} * D_{min} = 0$?
First, we assume that $A$ and $D$ are diagonalizable. Thus, let $S$ be a matrix with columns consisting of eigenvectors of $A$ so that $S^{-1} A S$ has diagonal form, and similarly $T$ so that $T^{-1} D T$ is diagonal.
Then by the laws of matrix multiplication, one can easily see that $$R:=\left(\begin{matrix}S &0\\ 0& T\end{matrix}\right)$$ does the same for $M$, i.e. $R^{-1} MR$ has diagonal form.
The other implication works similarly: Given a matrix $R$ that diagonalizes $M$, it can be written as consisting of four quadratic matrices $$ \left(\begin{matrix}S & ? \\ ?& T\end{matrix}\right).$$ But after matrix-multiplication with $M$, one realizes that the $?$-blocks in $R$ actually correspond to the $0$-blocks in $M$, so we can forget about them and decompose $R$ into the two matrices $S$ and $T$ that diagonalize $A$ and $D$.